0-1背包问题

真的好难 

描述

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N(1 ≤ N≤ 3,402) available charms. Each charm iin the supplied list has a weight W_i(1 ≤ W_i≤ 400), a 'desirability' factor D_i(1 ≤ D_i≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M(1 ≤ M≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

输入

Line 1: Two space-separated integers: N and M
Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

输出

Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

样例输入

4 6
1 4
2 6
3 12
2 7

样例输出

23

来源

USACO 2007 December Silver

按照正确的思路写了好长时间，最后一直得不到正确的结果，可能是状态转移方程没有理解清楚，而且还涉及到空间优化的问题（人人为我）

状态转移方程真的不太好理解，方程一直在变根本就找不到它确切的值。这里先贴上别人的AC代码，以后方便查找

学0-1背包前要看的一道题

点击打开链接（神奇的口袋）

关于0-1背包的讲解

点击打开链接

点击打开链接

空间可以优化

C/C++ (clang++ 3.3)

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#include<iostream>  

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#include<fstream> 

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#include<string.h> 

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using namespace std;  

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static const int N = 3403;  

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static const int M = 12881;  

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static int f[N][M];  		//表示在前N个物品中取重量不超过M的最佳价值

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static int c[N], v[N];  

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int main()  

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{  

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    int n, m;  			//n是物品的数量，m是背包的总容量

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    while (cin >> n >> m)  

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    {  

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        int i, j;  

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        for (i = 1; i <= n; i++)  

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        {  

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            cin >> c[i] >> v[i];  //c[i]重量	v[i]价值

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        }  

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        memset(f, 0, sizeof(f));  

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        for (i = 1; i <= n; i++)  

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        {  

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            for (j = 1; j < c[i]; j++)  //背包容量小于物体重量 ->不放

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                f[i][j] = f[i - 1][j];  //价值与上一个状态的价值一样

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            for(j = c[i]; j <= m; j++) 

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            {  

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                if (f[i - 1][j] < f[i - 1][j - c[i]] + v[i])  	//i和j参数的变化控制状态变化

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                    f[i][j] = f[i - 1][j - c[i]] + v[i];  

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                else  

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                    f[i][j] = f[i - 1][j]; 

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            }  

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        }  

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        cout << f[n][m] << "\n";  

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    }  

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    return 0;  

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}

空间优化后的代码

C/C++ (clang++ 3.3)

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#include<iostream>

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#include<string.h>

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#define M 12881

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#define N 3403

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​

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using namespace std;

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​

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int f[M];

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int w[N],v[N];

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​

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int main()

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{

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    int n,m;

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    cin>>n>>m;

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    for(int k=1; k<=n; k++)

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    {

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        cin>>w[k]>>v[k];

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    }

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​

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    memset(f, 0, sizeof(f)); 

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    for(int i=1; i<=n; i++)

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    {

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        for(int j=m; j>=w[i]; j--)

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        {

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            if(f[j]<f[j-w[i]]+v[i])

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                f[j]=f[j-w[i]]+v[i];

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        }

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    }

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    cout<<f[m]<<"\n";

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    return 0;

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}

代码虐我千百遍，我待代码如初恋。